why the multiplication between functionals is not well defined? I'm reading Olver (2012), chapter 7, and it says that the multiplication between functionals is not well defined. Then he refers to the section 5.4 where gives an explanation why, but I can't understand it.
Why I'm doing this question? well, I'm wondering why the notion of a Poisson bracket on the finite-dimensional setting drop off the Leibniz rule, see Kolev (2007). In section 3.3 when is defining a Hamiltonian Structure(A Poisson Structure but in functional spaces) it says that there is no well defined multiplication in functional spaces. 
Somebody understand why the Hamiltonian structures(as Poisson Structures on infinite-dimensinal setting) drop out the Leibniz rule?
I'm wondering if it has something to do with "Sur l’impossibilité de la multiplication des distributions" but I'm not sure.
 A: In this context "the" multiplication refers to a naive one [Olver, §5.4, p. 356]:

For example, the differential functions $u_x$ and $u_{xxx}$ both determine trivial functionals: $\int u_x dx = 0 = \int u_{xxx} dx$, but their product $u_xu_{xxx}$ is not a divergence, and hence $\mathscr{L} = \int u_xu_{xxx} dx$ is not a trivial functional. Indeed, $\delta \mathscr{L} = -2u_{xxxx} \neq 0$, hence by Theorem 4.7, $\mathscr{L} \neq 0$.

Here it is used that $\int D_x(\forall) dx = 0$ (boundary terms are assumed to vanish in this section), $u_x = D_x(u)$, $u_{xxx} = D_x(u_{xx})$, and $\delta \int D_x(\forall)dx = 0$.
The space of functionals $\mathscr{F}$ is defined as the space of differential functions modulo total divergences [loc. cit.], and the quotient map is denoted $L \mapsto \int L dx$. The above shows that there is no induced multiplication on $\mathscr{F}$ because it would depend on the choice of representatives: if such multiplication $\cdot$ were defined, then on the one hand $\int 0 dx \cdot \int 0 dx = \int 0 dx$ by direct computation, but on the other hand $\int 0 dx \cdot \int 0 dx \neq \int 0 dx$ by the example above; this is a contradiction.
Since there is no (such) multiplication, there is no Leibniz rule with respect to it. As noted in [Olver, §7.1 p. 435]:

However, the principal use of Leibniz' rule was to deduce the existence of a Hamiltonian vector field from a real-valued function $H$. This does carry over to the functional case:

